【LOJ115】无源汇有上下界可行流 模板题

Description

https://loj.ac/problem/115

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Solution

如果$u$向$v$有一条下界为$f_1$、上界为$f_2$的边，那么从$S$向$v$、$u$向$T$连一条流量为$f_1$的边，从$u$向$v$连一条$f_2-f_1$的边。当最大流等于$\sum f_1$时，有可行流。

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Code

跑得好慢哇，LOJ上都排到倒数第二版了。

/************************************************
 * Au: Hany01
 * Date: Jul 19th, 2018
 * Prob: 无源汇上下界可行流
 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
	static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 205, maxm = 10205 * 6;

int n, m, S, T, v[maxm], nex[maxm], beg[maxn], e = 1, d[maxn], gap[maxn], f[maxm], Ans, tot;

struct Edge { int u, v, f1, f2; }E[10205];

inline void add(int uu, int vv, int ff, int mk = 1) {
	v[++ e] = vv, f[e] = ff, nex[e] = beg[uu], beg[uu] = e;
	if (mk) add(vv, uu, 0, 0);
}

int sap(int u, int flow)
{
	if (u == T) return flow;
	register int res = flow, tmp;
	for (register int i = beg[u]; i; i = nex[i]) {
		if (d[u] != d[v[i]] + 1 || !f[i]) continue;
		tmp = sap(v[i], min(res, f[i]));
		f[i] -= tmp; f[i ^ 1] += tmp;
		if (!(res -= tmp)) return flow;
	}
	if (!--gap[d[u]]) d[S] = T;
	++ gap[++ d[u]];
	return flow - res;
}

int main()
{
#ifdef hany01
	File("loj115");
#endif

	n = read(), m = read(), T = (S = n + 1) + 1;
	For(i, 1, m)
		E[i].u = read(), E[i].v = read(), tot += E[i].f1 = read(), E[i].f2 = read(),
		add(S, E[i].v, E[i].f1), add(E[i].u, E[i].v, E[i].f2 - E[i].f1), add(E[i].u, T, E[i].f1);

	for (gap[0] = T; d[S] < T; ) Ans += sap(S, INF);
	if (Ans < tot) { puts("NO"); return 0; }
	puts("YES");
	For(i, 1, m) printf("%d\n", f[(i - 1) * 6 + 5] + E[i].f1);

    return 0;
}